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Thread: second order taylor polynomial

  1. #1
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    Question second order taylor polynomial

    Hi i was wondering if anyone could help me with this question.

    i need to use the second order taylor polynomial with remainder to estimate sqraure-root of 101 and give an estimate of the error mode.

    but where i get confused is that i am using the formula for the taylor poly..and i dont know what f(a) so i cant dfferentiate and so i cant fill in the taylor poly formula..all i am is given is square root of 101..

    can someone please help thanks
    Last edited by dopi; Nov 4th 2007 at 10:24 AM.
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  2. #2
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    Quote Originally Posted by dopi View Post
    Hi i was wondering if anyone could help me with this question.

    i need to use the second order taylor polynomial with remainder to estimate sqraure-root of 101 and give an estimate of the error mode.

    but where i get confused is that i am using the formula for the taylor poly..and i dont know what f(a) so i cant dfferentiate and so i cant fill in the taylor poly formula..all i am is given is square root of 101..

    can someone please help thanks
    Let $\displaystyle f(x) = \sqrt{x+100} = (x+100)^{1/2}$ the Taylor polynomial centered at zero or order 2 is, $\displaystyle T_2(x) = 10 + \frac{x}{20} - \frac{x^2}{4000}$ consider the interval $\displaystyle (-2,2)$ the derivative is $\displaystyle f'(x) = \frac{1}{2}(x+100)^{-1/2}$ which is bounded by $\displaystyle \frac{1}{2}(98)^{1/2}$.
    So the maximum error possible,
    $\displaystyle \frac{\sqrt{98}}{2\cdot 3!}$
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  3. #3
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    Hello I have the same problem and I cannot see how you estimated the square root (101)
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  4. #4
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    Quote Originally Posted by varkoume.com View Post
    Hello I have the same problem and I cannot see how you estimated the square root (101)
    The exact value is $\displaystyle f(1)$ and the estimate is $\displaystyle T_2 (1)$.
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